This is my solution not sure if it is correct though.
Solution: For the expression to be an integer, $(n-4)\sqrt{n-19}$ should be a perfect square. For $(n-4)\sqrt{n-19}$ to be a perfect square, $(n-4)^2(n-19)$ should be a perfect square.
Now consider some perfect square $k^2$ where $k\in\Bbb{Z}$. If I multiply it by some integer $r$ the perfect square will no longer be a perfect square unless $r=1,r=0,k=0$.
We see that above, $(n-4)^2$ is a perfect square and $n-19$ is some integer. So for $(n-4)^2(n-19)$ to be a perfect square, $(n-4)^2$ must be zero or $(n-19)$ must be zero or one.
As a result there are three possible $n$'s such that the expression is an integer, $n=4,n=19,n=20$.
Edit: This solution is wrong!
If you start with $$n=s^2+19$$ and substitute that in and just follow the algebra you get 20 and 244. $$\sqrt{(n-4)\sqrt{n-19}}=\sqrt{(s^2+15)s}=q$$ Then$$(s^2+15)s=q^2$$ $$s^3+15 s=q^2$$ Then this means that $$s|q$$ so $$q=s t$$ Then you get $$s^2+15=s t^2$$ by which you infer that $$s|15$$ You then try all the factors of 15 and get 20 or 244.